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displayed the main inequality
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edit approved Jun 2, 2015 at 14:54
Iosif Pinelis
edit approved Jun 2, 2015 at 14:54
Iosif Pinelis
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I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$.

$$(*)\qquad N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|.$$

Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that

$$(*)\qquad N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|.$$

Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

added 6 characters in body
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fedja
fedja
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I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}(e_i+e_j)$$v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

added 7 characters in body
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fedja
fedja
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I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ linearly independentstandard basis vectors $e_n$ of the same length and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\le N(N-1)\|e_1-e_2\|$$N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i\ne j}|a_{i,j}|$$\sum_{i< j}|a_{i,j}|$ with $a_{ij}=a_{ji}$ and $v=\sum_{i\ne j}(e_i+e_j)$ $v=\sum_{i< j}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i\ne j}a_{i,j}(e_i+e_j)\,, $$$$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=2\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i,j}|a_{ij}|$$2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ linearly independent vectors $e_n$ of the same length and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\le N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i\ne j}|a_{i,j}|$ with $a_{ij}=a_{ji}$ and $v=\sum_{i\ne j}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i\ne j}a_{i,j}(e_i+e_j)\,, $$ then $2=2\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i,j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The answer is "No".

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that $N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|$. Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$ e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,, $$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

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fedja
fedja
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  • 160
  • 302